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Problem solution · C++

Subarrays Distinct Element Sum of Squares II

Subarrays Distinct Element Sum of Squares II: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Segment tree or range structure
Source
walkccc LeetCode Solutions
Length
73 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Segment tree or range structure

For Subarrays Distinct Element Sum of Squares II, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 73 lines of C++ from the credited upstream file 2916.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 1 loop block detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSubarrays Distinct Element Sum of Squares II · C++C++
Use this to learn the idea, then write your own version.
class LazySegmentTree { public:  LazySegmentTree(int n, int kMod)      : n(n), kMod(kMod), lazy(4 * n), sums(4 * n), squaredSums(4 * n) {}   void updateRange(int l, int r) {    return updateRange(/*i=*/0, /*start=*/0, /*end=*/n - 1, l, r);  }   void propagate(int i, int l, int r) {    const int gap = r - l + 1;    // (a + L)^2 + (b + L)^2 + (c + L)^2, where L = lazy[i]    // a^2 + b^2 + c^2 + 2 * L (a + b + c) + L^2 * gap, where gap = 3    squaredSums[i] += 2 * lazy[i] * sums[i] + lazy[i] * lazy[i] * gap;    squaredSums[i] %= kMod;    sums[i] += lazy[i] * gap;    sums[i] %= kMod;    if (l < r) {      lazy[i * 2 + 1] += lazy[i];      lazy[i * 2 + 2] += lazy[i];    }    lazy[i] = 0;  }   int getTreeSquaredSums() {    return squaredSums[0];  }  private:  const int kMod;  const int n;  vector<long> lazy;  vector<long> sums;  vector<long> squaredSums;   void updateRange(int i, int start, int end, int l, int r) {    if (lazy[i] > 0)      propagate(i, start, end);    if (end < l || start > r)      return;    if (start >= l && end <= r) {      lazy[i] = 1;      propagate(i, start, end);      return;    }    const int mid = (start + end) / 2;    updateRange(i * 2 + 1, start, mid, l, r);    updateRange(i * 2 + 2, mid + 1, end, l, r);    sums[i] = (sums[i * 2 + 1] + sums[i * 2 + 2]) % kMod;    squaredSums[i] = (squaredSums[i * 2 + 1] + squaredSums[i * 2 + 2]) % kMod;  }}; class Solution { public:  int sumCounts(vector<int>& nums) {    constexpr int kMod = 1'000'000'007;    const int n = nums.size();    int ans = 0;    unordered_map<int, int> lastSeen;    LazySegmentTree tree(n, kMod);     for (int r = 0; r < n; ++r) {      const int l = lastSeen.contains(nums[r]) ? lastSeen[nums[r]] + 1 : 0;      tree.updateRange(l, r);      lastSeen[nums[r]] = r;      ans = (ans + tree.getTreeSquaredSums()) % kMod;    }     return ans;  }};

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